Asking this because I'm not sure how "mathematically valid" is the argument that I used in my demonstration. Please note that my question is not about how to prove the theorem, but about whether my demonstration is correct. If the proof is incorrect, is there any way to make it more "valid mathematically"? If it's correct, my question is: what can be done to improve it?
Theorem. Let $E$ be a finite vector space with dimension $N$.
Every linearly independent set $X=\{v_1, ..., v_m\}$ of $E$ is a subset of a basis for $E$.
Proof.
Firstly, note that $m<N$. Due to the hypothesis, we know that $\exists v\in E-Span(\{v_1,...,v_m\})$. From here, there are two possibilities:
(A) $Span(\{v_1,...,v_m,v\})=E$
(B) There still exists another $\exists v\in E - Span(\{v_1, ..., v_m, v\})$
In either scenario, $\{v_1,...,v_m, v\}$ is linearly independent. If (A) is true, then we are done. In the case of (B), we repeat the process with $\{v_1,...,v_m,v\}$ instead of $\{v_1,...,v_m\}$ until we get (A).
(Part that I don't know whether it is necessary or not:)
We claim that the process described happens in a finite number of times. In fact, there's no way for it to happen more times than $N$ times. If the process were to be repeated, let's say $N+1$ times, then we would obtain a linearly independent set that has a cardinality greater than $\dim{E}$, which is absurd.
It is correct, but you should never use the same symbol for two distinct objects. I'm talking about $v$ here.
There is a problem with your notation: you should have written $E\setminus\operatorname{Span}(\{e_1,\ldots,e_n\})$ instead of $E-\operatorname{Span}(\{e_1,\ldots,e_n\})$ (it is ambiguous).
Finally, it would be more elegant to express your proof as an inductive proof on $N-m$.